Does the heating of a wire depend on the frequency of current flowing through it? Does heating of a wire only depend on how much current is flowing through it, or does it also depend on the frequency of that current (if it is AC)? If so, what is the relationship between frequency and heating?
Thanks!
 A: Heating of the wire can indeed significantly depend on the frequency of the current due to the skin effect, among other things. Skin effect causes an increase of the effective resistance of the wire as the current flows closer to the periphery of the wire at higher frequency.
A: Yes the heating of a metal or resistor depends on frequency. There are many types of resistors. Metal oxide, metal foil, high power (2W ~ 10W), etc...
All resistor have properties like temperature coefficient of resistance (TCR), which is gives an idea of how the resistor value will change with the temperature conditions. Usually, the analog resistor is placed at room temperature. So in that case, you might think that resistor with good TCR will not change with the room temperature, that is right. But it can changes according to the AC input power on this resistor.
Usually, TCR is about 0.01 ppm/K to typically 100 ppm/K. If you flow a AC current on a resistor $I=I_o \sin(\omega t)$ , due to Corbino effect, you get a third harmonic voltage V3w that contains informations on how the wire is heated and it depends on the frequency.
$R = R_o[1+a(T-T_o)]$. The power is $P = RI^2$ at $2\omega$ so the elevation of temperature on the resistor is also at $2\omega$ (harmonic 2). Now the voltage is $V = RI$, which is the product of a 2w signal with $1\omega$ signal, like $\sin(2\omega t)\sin(\omega t)$, so you get a $3\omega$ signal that you can detect with a lock in amplifier, with the reference to the input 1w signal of the current source.  Therefore you can monitor the heating of a wire as a function of frequency, by detecting the $3\omega$ (third harmonic) of your signal with a lock in amplifier.
A: Due to the skin effect heating (power $P$ dissipated in the wire) depends on frequency, but not in a simple or linear way. As long as the skin depth is larger that the radius of the wire, you see almost no effect. When the skin depth gets smaller than about 50% of the radius, but not being super small, $P(f)$ is proportional to $\sqrt{f}$. At super small skin depths $P(f)$ should be proportional to $f$.
A: Ideally speaking, yes and no.  If you mean the average, no it wouldn't change with the frequency, but if you want the instantaneous heat,  then yes it totally depends on the frequency of the current. But in the real world you're usually concerned with the average, so it wouldn't matter.
MATH
Ideally speaking, a wire is not supposed to drop a voltage, but the real world is more fickle than a cat, so let's say it drops a small voltage, ie, let's model it as a resistor.
Then the equation would be-
$$VI = P$$
$$V = V_o\cos(\omega t)$$
$$I = \frac{V_o}{R}\cos(\omega t)$$
Simply multiply to get the answer
$$P = \frac{1}{R}V_o^2\cos^2(\omega t)$$
This, you see, is the instantaneous power, but when we're in the real world, instantaneous power is pretty much useless in this case, so we use the average power-
$$P_{avg} = \frac{1}{T}\int^\pi_0 P.d(\omega t)$$
And since I'm too lazy, I won't do much further, but what you should know is that it's constant. (It doesn't contain $\omega$) So it doesn't depend on the frequency
